id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
values | problem_type stringclasses 4
values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
01fa | A fair $k$-die is a die with $k$ sides and the numbers $1, 2, \dots, k$ written one on each side such that the probability of getting one of these numbers when rolling the die is exactly $\frac{1}{k}$ for each number. Let $n$ be a positive integer and $a_1 \ge a_2 \ge \dots \ge a_n > 0$ be integers. Georg has $n$ fair ... | [] | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
01ch | Prove that, for positive $x$, $y$, $z$, the following inequality holds:
$$
(x + y + z)(4x + y + 2z)(2x + y + 8z) \geq \frac{375}{2}xyz.
$$ | [
"Consider the first two brackets and observe that\n$$\n(x + y + z)(4x + y + 2z) = (2x + y)^2 + 3z(2x + y) + 2z^2 + xy.\n$$\nTherefore, we can write the inequality in the form\n$$\n\\left( \\frac{(2x + y)^2 + 3z(2x + y) + 2z^2}{xy} + 1 \\right) \\cdot \\frac{2x + y + 8z}{z} \\geq \\frac{375}{2}.\n$$\nNow fix $z$ and... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
03xw | It is known that $\{a_n\}$ is an arithmetic sequence with non-zero common difference and $\{b_n\}$ a geometric sequence, satisfying $a_1 = 3$, $b_1 = 1$, $a_2 = b_2$, $3a_5 = b_3$; furthermore, there are constants $\alpha$ and $\beta$ such that for every positive integer $n$, we have $a_n = \log_a b_n + \beta$. Then $\... | [
"Let the common difference of $\\{a_n\\}$ be $d$ and the common ratio of $\\{b_n\\}$ be $q$. Then\n$$\n3 + d = q, \\qquad \\textcircled{1}\n$$\n$$\n3(3 + 4d) = q^2. \\qquad \\textcircled{2}\n$$\nSubstituting $\\textcircled{1}$ into $\\textcircled{2}$, we have $9 + 12d = d^2 + 6d + 9$. Then we get $d = 6$ and $q = 9... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | final answer only | 3^(1/3) + 3 | |
0jl2 | Problem:
We have two concentric circles $C_{1}$ and $C_{2}$ with radii $1$ and $2$, respectively. A random chord of $C_{2}$ is chosen. What is the probability that it intersects $C_{1}$?
Your answer to this problem must be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are positive integers. If your answer is... | [] | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Circles"
] | null | final answer only | 1/3 | |
0f9e | Problem:
$A$, $B$, $C$ are adjacent vertices of a regular $2n$-gon and $D$ is the vertex opposite to $B$ (so that $BD$ passes through the center of the $2n$-gon). $X$ is a point on the side $AB$ and $Y$ is a point on the side $BC$ so that $\angle XDY = \pi/2n$. Show that $DY$ bisects $\angle XYC$. | [] | Soviet Union | 24th ASU | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
07uo | A circle, centre $I$, lies inside a circle, centre $J$, and touches it at a point $A$. A tangent is drawn to the circle, centre $I$, at a point $B$ different from $A$ which intersects the circle, centre $J$, at $C$ and $D$. Prove that $AB$ bisects $\angle CAD$. | [
"Let $T$ be the point on the line $CD$ for which $TA$ is the common tangent to both circles. Then $|TA| = |TB|$, because $TB$ is a tangent to the smaller circle as well. Hence $\\angle TBA = \\angle TAB$.\n\n\n\nLooking at triangle $ABC$ we see that $\\angle TBA = \\angle BDA + \\angle BAD$... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08fq | Problem:
Le caselle di una scacchiera $9 \times 10$ sono colorate di rosso, verde o blu. Ogni quadrato $3 \times 3$ della scacchiera contiene esattamente tre caselle rosse, tre verdi e tre blu. Quante sono al massimo le caselle rosse?
(A) 32
(B) 34
(C) 36
(D) 39
(E) 40 | [
"Solution:\nLa risposta è $\\mathbf{( C )}$. Consideriamo un quadrato $9 \\times 9$ della scacchiera, come in figura. Questa sotto-scacchiera può essere coperta con $\\frac{9 \\times 9}{9}=9$ quadrati $3 \\times 3$, ognuno dei quali contiene per ipotesi 3 caselle rosse. Le caselle rosse sono quindi al massimo pari ... | Italy | Italian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | C | |
03qf | Let $p$ be an odd prime. Let $k$ be a positive integer such that $\sqrt{k^2 - pk}$ is also a positive integer. Then $k = \underline{\hspace{2cm}}$. | [
"Set $\\sqrt{k^2 - pk} = n$, $n \\in \\mathbb{N}$. Thus $k^2 - pk - n^2 = 0$, and $k = \\frac{p \\pm \\sqrt{p^2 + 4n^2}}{2}$, which implies that $p^2 + 4n^2$ is a perfect square, say $m^2$, where $m \\in \\mathbb{N}$. So $(m-2n)(m+2n) = p^2$.\n\nSince $p$ is a prime and $p \\ge 3$, we have\n$$\n\\begin{cases} m - 2... | China | China Mathematical Competition (Hainan) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (p+1)^2/4 | |
043h | In the plane rectangular coordinate system, the graph of function $y = \frac{x+1}{|x|+1}$ has three different points lying on line $l$, and the sum of the abscissas of these three points is $0$. Find the range of values of the slope of $l$. | [
"When $x \\ge 0$, $y = 1$; when $x < 0$, $y = \\frac{x+1}{1-x}$ is strictly increasing about $x$ and less than $1$.\n\nSuppose line $l: y = kx + b$, and then the known conditions are equivalent to the fact that equation\n$$\nkx + b = \\frac{x+1}{|x|+1} \\qquad (1)\n$$\nhas three different real number solutions $x_1... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 0 < k < 2/9 | |
0a91 | Problem:
A sequence of positive integers $\{a_{n}\}$ is given by
$$
a_{0}=m \quad \text{ and } \quad a_{n+1}=a_{n}^{5}+487
$$
for all $n \geq 0$. Determine all values of $m$ for which the sequence contains as many square numbers as possible. | [
"Solution:\nConsider the expression $x^{5}+487$ modulo $4$. Clearly $x \\equiv 0 \\Rightarrow x^{5}+487 \\equiv 3$, $x \\equiv 1 \\Rightarrow x^{5}+487 \\equiv 0$; $x \\equiv 2 \\Rightarrow x^{5}+487 \\equiv 3$, and $x \\equiv 3 \\Rightarrow x^{5}+487 \\equiv 2$.\n\nSquare numbers are always $\\equiv 0$ or $\\equiv... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 20 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 9 | |
0f0q | Problem:
Define $p(x) = a x^2 + b x + c$. If $p(x) = x$ has no real roots, prove that $p(p(x)) = 0$ has no real roots. | [] | Soviet Union | ASU | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
099v | A certain factory produces asphalt pavements in the shape of right hexagons with unit edges and unit width. During transportation some corners of the pavement may incur some degree of damages. Then can 7 pavements with the same degree damage be found at all times from 2009 pavements?
(proposed by B. Bayasgalan) | [
"There are 12 permutations of this object that transform it onto itself that are unit and in the form of $(1\\ 2\\ 3\\ 4\\ 5\\ 6)$; $(7\\ 8\\ 9\\ 10\\ 11\\ 12)$; $(1\\ 3\\ 5)(2\\ 4\\ 6)(7\\ 9\\ 11)(8\\ 10\\ 12)$; $(1\\ 4)(3\\ 6)(2\\ 5)(3\\ 6)(7\\ 10)(8\\ 10)(9\\ 12)$; $(1\\ 5\\ 3)(2\\ 6\\ 4)(7\\ 11\\ 9)(8\\ 12\\ 10... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | No | |
03sy | Assume that $a$ is a positive integer and not a perfect square. Prove that for any positive integer $n$, the sum
$$
S_n = \{\sqrt{a}\} + \{\sqrt{a}\}^2 + \dots + \{\sqrt{a}\}^n
$$
is irrational, where $\{x\} = x - \lfloor x \rfloor$ and $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. | [
"Suppose that $c^2 < a < (c+1)^2$, where $c$ is an integer greater than or equal to $1$. Then, $\\lfloor \\sqrt{a} \\rfloor = c$, $1 \\le a - c^2 \\le 2c$, and $\\{\\sqrt{a}\\} = \\sqrt{a} - c$.\n\nWrite $\\{\\sqrt{a}\\}^k = (\\sqrt{a} - c)^k = x_k + y_k \\sqrt{a}$, where $k \\in \\mathbb{N}$ and $x_k, y_k \\in \\m... | China | China Western Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
048l | Two players play the following game: after one of them tells number $n$, the other has to tell a number of the form $a \cdot b$ where $a, b$ are positive integers such that $a + b = n$. The game continues in the same way. If at some point one of the players has told $2011$, which are the possible numbers that the game ... | [] | Croatia | Hrvatska 2011 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All integers n ≥ 5 | |
0bvy | Triangle $ABC$ has $\hat{B}, \hat{C} < 90^\circ$ and $AB \neq AC$. Denote $m, n, p$ the measures of the angles made by the altitude, the bisector, respectively the median from $A$ with $BC$. Show that the angle $\hat{A}$ is right if and only if $n$ is the arithmetic mean of $m$ and $p$.
Claudiu Ştefan Popa | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | English | proof only | null | |
02td | Problem:
João trabalha vendendo pacotes de previsão astrológica. Para incrementar as vendas de suas previsões, ele oferece descontos caso pessoas de um mesmo signo queiram contratar seus serviços. No Horóscopo Grego, como existem exatamente 12 signos, portanto, em um grupo de 13 pessoas, sempre duas delas terão o mesm... | [
"Solution:\n\na) O mínimo é 25. Se em um grupo de 24 pessoas cada signo aparecer no máximo duas vezes, teremos no máximo $2 \\cdot 12=24$ pessoas. Como $24<25$, isso mostra que pelo menos um dos signos deverá aparecer três vezes. De fato, esse é o mínimo onde tal propriedade ocorre pois se considerarmos 24 pessoas ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | a) 25; b) 145 | |
0fks | Problem:
Dada una circunferencia y en ella dos puntos fijos $A$, $B$, otro variable $P$ y una recta $r$; se trazan las rectas $PA$ y $PB$ que cortan a $r$ en $C$ y $D$ respectivamente. Determina dos puntos fijos de $r$, $M$ y $N$, tales que el producto $CM \cdot DN$ sea constante al variar $P$. | [
"Solution:\n\nTrazamos las paralelas a $r$ por $A$ y $B$ que cortan a la circunferencia en $A'$ y $B'$ respectivamente de modo que $AA'BB'$ es un trapecio isósceles.\nLas intersecciones de $AB'$ y $BA'$ con $r$ determinan los puntos $M$ y $N$ buscados.\n\nEn efecto, los triángulos $AMC$ y $DNB$ (sombreados en la fi... | Spain | XLIV Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Draw the lines through each fixed point parallel to the given line; let their intersections with the circle be A′ and B′. Let M be the intersection of the given line with the line through the first fixed point and B′, and let N be the intersection of the given line with the line through the second fixed point and A′. T... | |
0jon | Problem:
$ABCD$ is a cyclic quadrilateral with sides $AB = 10$, $BC = 8$, $CD = 25$, and $DA = 12$. A circle $\omega$ is tangent to segments $DA$, $AB$, and $BC$. Find the radius of $\omega$. | [
"Solution:\n\nDenote $E$ as the intersection point of $AD$ and $BC$. Let $x = EA$ and $y = EB$. Because $ABCD$ is a cyclic quadrilateral, $\\triangle EAB$ is similar to $\\triangle ECD$. Therefore,\n$$\n\\frac{y + 8}{x} = \\frac{25}{10} \\quad \\text{and} \\quad \\frac{x + 12}{y} = \\frac{25}{10}.\n$$\nWe get $x = ... | United States | HMMT February | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | sqrt(8463)/7 | |
05eq | Problem:
Soit $a$, $b$ et $c$ des réels tels que
$$
|a-b| \geq |c|,\ |b-c| \geq |a| \text{ et } |c-a| \geq |b|.
$$
Prouver que l'un des trois nombres $a$, $b$ et $c$ est la somme des deux autres. | [
"Solution:\nLa première inégalité s'écrit $(a-b)^2 \\geq c^2$, ou encore $(a-b+c)(a-b-c) \\geq 0$. De même, les deux autres inégalités conduisent à $(b-c+a)(b-c-a) \\geq 0$ et $(c-a+b)(c-a-b) \\geq 0$.\nEn multipliant membre à membre ces trois inégalités, il vient\n$$\n(a-b+c)^2(-a+b+c)^2(a+b-c)^2 \\leq 0\n$$\nOr, ... | France | null | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0h8m | Solve the system of equations for positive integer numbers $x, y, z$:
$$
\begin{cases} x^3 - 6y^2 + 27z = 132, \\ y^3 - 9z^2 + 3x = 125, \\ z^3 - 3x^2 + 12y = -68. \end{cases}
$$ | [
"Let's add up all three equations:\n\n$$\n(x^3 - 3x^2 + 3x) + (y^3 - 6y^2 + 12y) + (z^3 - 9z^2 + 27z) = 189 \\Leftrightarrow \\\\\n(x^3 - 3x^2 + 3x - 1) + (y^3 - 6y^2 + 3y - 27) + (z^3 - 9z^2 + 27z - 27) = 153 \\Leftrightarrow \\\\\n(x-1)^3 + (y-2)^3 + (z-3)^3 = 153.\n$$\n\nBy a simple exhaustive search, we can see... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | No positive integer solutions | |
0eov | $\triangle ABE$ and $\triangle BCF$ are equilateral triangles and $ABCD$ is a square.
Prove that $\triangle DEF$ is an equilateral triangle.
 | [
"First note that reflecting the figure around the line $BD$ doesn't change the figure (the equilateral triangle $AEB$ reflects to the triangle $CBF$). This means that $DE$ and $DF$ are symmetric about the line $BD$, which means that $DF = DE$. If we can show that $\\angle FDE = 60^\\circ$, then triangle $DEF$ is an... | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0buq | Problem:
Fie $a \in (0,4)$, a fixat. Determinați $z$ din $\mathbb{C}$ știind că $|z| = \left|\frac{a}{z}\right| = |z - a|$. | [] | Romania | Olimpiada de Matematică Etapa Locală | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof and answer | z = a/2 ± (i/2)·sqrt(a(4 − a)) | |
0bio | In a circle, consider two chords $[AB]$, $[CD]$ that intersect at $E$. The lines $AC$ and $BD$ meet at $F$. Let $G$ be the projection of $E$ onto $AC$. We denote by $M, N, K$ the midpoints of the segment lines $[EF]$, $[EA]$, and $[AD]$, respectively. Prove that the points $M, N, K, G$ are concyclic.
Marius Bocanu
![... | [
"Let $P$ be the midpoint of $[AF]$; points $N, G, P, M$ are on the Euler circle of triangle $AEF$, which means they are concyclic $(*)$.\n\nAs $NK \\parallel DE$ and $NM \\parallel AF$, we have:\n$$\n\\begin{align*}\n\\angle KNM &= \\angle KNE + \\angle ENM = \\angle DEB + \\angle BAC \\\\\n&= \\angle DEB + \\angle... | Romania | 65th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07nu | Find the least positive integer $a$ such that $2013$ divides $94^n + a \cdot 28^n$ for some positive integer $n$. | [
"Because $2013 = 3 \\cdot 11 \\cdot 61$, the requirement $94^n + a \\cdot 28^n \\equiv 0 \\pmod{2013}$ implies $a+1 \\equiv 0 \\pmod{3}$, $6^n(a+1) \\equiv 0 \\pmod{11}$ and $28^n((-1)^n + a) \\equiv 0 \\pmod{61}$. Therefore $a \\equiv -1 \\pmod{3}$, $a \\equiv -1 \\pmod{11}$ and $a \\equiv (-1)^{n+1} \\pmod{61}$. ... | Ireland | Ireland | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 428 | |
02w3 | Problem:
Qual a maior quantidade de inteiros que podemos escolher no conjunto $\{1,2,3, \ldots, 2017\}$ de modo que a diferença entre quaisquer dois deles não seja um número primo? | [
"Solution:\n\nA maior quantidade é $\\frac{2017-1}{4}+1=505$, que pode ser obtida escolhendo-se os elementos que deixam resto 1 na divisão por 4: $\\{1,5,9, \\ldots, 2017\\}$. A diferença entre quaisquer dois deles é sempre um múltiplo de 4 e consequentemente não é um número primo. Mostraremos agora que não é possí... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 505 | |
0b14 | Problem:
Let $N = 2019^{2} - 1$. How many positive factors of $N^{2}$ do not divide $N$? | [
"Solution:\nNote that $N = 2^{3} \\cdot 5 \\cdot 101 \\cdot 1009$ and so $N$ has $4 \\cdot 2 \\cdot 2 \\cdot 2 = 32$ factors. On the other hand, $N^{2} = 2^{6} \\cdot 5^{2} \\cdot 101^{2} \\cdot 1009^{2}$ and so $N^{2}$ has $7 \\cdot 3 \\cdot 3 \\cdot 3 = 189$ factors. Hence, $189 - 32 = 157$ of these factors of $N... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 157 | |
0857 | Problem:
Si consideri la disuguaglianza
$$
\left(x_{1}+\ldots+x_{n}\right)^{2} \geq 4\left(x_{1} x_{2}+x_{2} x_{3}+\ldots+x_{n} x_{1}\right) .
$$
a. Determinare per quali $n \geq 3$ è vera per ogni possibile scelta di numeri reali positivi $x_{1}, \ldots, x_{n}$.
b. Determinare per quali $n \geq 3$ è vera per ogni p... | [
"Solution:\n\nDimostreremo che la disuguaglianza\n- è vera per ogni scelta $x_{1}, \\ldots, x_{n}$ di numeri reali positivi se e solo se $n \\geq 4$;\n- è vera per ogni scelta $x_{1}, \\ldots, x_{n}$ di numeri reali se e solo se $n=4$.\n\nSuddividiamo la dimostrazione in vari passi.\n\nPasso 1. Dimostriamo che per ... | Italy | XXII OLIMPIADE ITALIANA DI MATEMATICA | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a: n ≥ 4; b: n = 4 | |
09am | Given are $m$ positive integers such that $1 \le a_1 \le a_2 \le \dots \le a_m \le 400$. Show that $m \le 40$, if $\text{gcd}(a_i, a_j) \le 400$ for all $i$ and $j$ ($1 \le i, j \le m$). | [
"First let us change the order: $a_1 = b_m$, $a_2 = b_{m-1}$, ..., $a_m = b_1$. Then we have $1 \\le b_m \\le b_{m-1} \\le \\dots \\le b_2 \\le b_1 \\le 400$. Let us show that $b_k \\le \\frac{400}{k}$, $1 \\le k \\le m$. If $k=1$, it is true.\n\nFor the induction step, we need to prove that $b_{k+1} \\le \\frac{40... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
02sv | Problem:
Um prédio tem três escadas diferentes, todas começando na base do prédio e terminando no topo. Uma escada tem 104 degraus, outra tem 117 degraus, e a outra tem 156 degraus. Sempre que os degraus das três escadas estão na mesma altura, há um andar. Quantos andares tem o prédio? | [
"Solution:\n\nVamos chamar de $A$, $B$ e $C$ as três escadas, que têm 104, 117 e 156 degraus, respectivamente. Seja $a$ o número de degraus da escada $A$ entre cada dois andares, $b$ o número de degraus da escada $B$ entre cada dois andares, e $c$ o número de degraus da escada $C$ entre cada dois andares. Dividindo... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 13 | |
0d0i | Let $S$ be the set of positive integers. Determine all functions $f: S \to S$ such that $x^2 + f(y)$ divides $f(x)^2 + y$ for every pair of positive integers $x$ and $y$. | [
"Consider $x = y = 1$. Then we get that $\\frac{1 + f^2(1)}{1 + f(1)}$ is an integer, so $1 + f(1)$ divides $2f(1)$. It follows $1 + f(1) = 2$, that is $f(1) = 1$.\n\nFor $y = 1$, we obtain $x^2 + 1$ divides $f^2(x) + 1$. Therefore $x^2 \\leq f^2(x)$, which means that $x \\leq f(x)$ for every positive integer $x$.\... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | f(x) = x for all positive integers x | |
0kw4 | Problem:
Compute
$$
\sum_{\substack{a+b+c=12 \\ a \geq 6,\ b,\ c \geq 0}} \frac{a!}{b!c!(a-b-c)!}
$$
where the sum runs over all triples of nonnegative integers $(a, b, c)$ such that $a+b+c=12$ and $a \geq 6$. | [
"Solution:\nWe tile a $1 \\times 12$ board with red $1 \\times 1$ pieces, blue $1 \\times 2$ pieces, and green $1 \\times 2$ pieces. Suppose we use $a$ total pieces, $b$ blue pieces, and $c$ green pieces. Then we must have $a+b+c=12$, and the number of ways to order the pieces is\n$$\n\\binom{a}{b, c, a-b-c} .\n$$\... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 2731 | |
0ecd | Problem:
V podjetju, ki ga vodi več direktorjev, imajo sef, ki je zaklenjen s šestimi ključavnicami. Vsak direktor ima tri ključe, s katerimi lahko odklene tri različne ključavnice. Z vsakim ključem lahko odklene natanko eno ključavnico.
Nobena dva direktorja ne moreta odkleniti istih treh ključavnic in nobena dva di... | [
"Solution:\n\n1. način. Podjetje vodi največ 10 direktorjev.\n\nOznačimo ključavnice oziroma pripadajoče ključe s številkami od 1 do 6. Vsak direktor ima set 3 ključev od 3 različnih ključavnic in nobena dva direktorja nimata enakih setov. Zato najprej preštejmo, koliko različnih setov 3 ključev obstaja. Za posamez... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 10 | |
0e6w | Let $ABCD$ be a square and $M$ the midpoint of the side $BC$. Let $P$ be the orthogonal projection of point $C$ onto the line segment $DM$. Prove that the triangle $DAP$ is an isosceles triangle with the top angle at $A$. | [
"Let $N$ be the midpoint of the side $CD$, and let $R$ be the intersection point of the lines $DP$ and $AN$. Then $\\angle DRA = 180^\\circ - \\angle ADR - \\angle NAD = 180^\\circ - \\angle ADR - \\angle MDC = 180^\\circ - 90^\\circ = 90^\\circ$. The line $RN$ is thus parallel to the line $PC$, and from $|DN| = |N... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0kw6 | Problem:
Let $\omega_{1}$ and $\omega_{2}$ be two non-intersecting circles. Suppose the following three conditions hold:
- The length of a common internal tangent of $\omega_{1}$ and $\omega_{2}$ is equal to $19$.
- The length of a common external tangent of $\omega_{1}$ and $\omega_{2}$ is equal to $37$.
- If two poi... | [
"Solution:\n\nThe key claim is that $\\mathbb{E}\\left[XY^{2}\\right] = d^{2} + r_{1}^{2} + r_{2}^{2}$.\n\nTo prove this claim, choose an arbitrary point $B$ on $\\omega_{2}$. Let $r_{1}, r_{2}$ be the radii of $\\omega_{1}, \\omega_{2}$ respectively, and $O_{1}, O_{2}$ be the centers of $\\omega_{1}, \\omega_{2}$ ... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 38 | |
0jj5 | Problem:
An auditorium has two rows of seats, with 50 seats in each row. 100 indistinguishable people sit in the seats one at a time, subject to the condition that each person, except for the first person to sit in each row, must sit to the left or right of an occupied seat, and no two people can sit in the same seat.... | [
"Solution:\n\nAnswer: $\\boxed{\\binom{100}{50} 2^{49}}$ First, note that there are $2^{49}$ ways a single row can be filled, because each of the 49 people after the first in a row must sit to the left or to the right of the current group of people in the row, so there are 2 possibilities for each of these 49 peopl... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | binom{100}{50} 2^{98} | |
055h | Let $ABC$ be an isosceles triangle with apex $A$ and altitude $AD$. On $AB$, choose a point $F$ distinct from $B$ such that $CF$ is tangent to the incircle of $ABD$. Suppose that $\triangle BCF$ is isosceles. Show that those conditions uniquely determine:
a) which vertex of $BCF$ is its apex;
b) the size of $\angle B... | [
"a) Consider cases of the location of the vertex angle of the triangle $BCF$ (Fig. 27).\nIf $F$ were the apex, then $F$ would lie on the perpendicular bisector of the side $BC$, i.e., on the line $AD$, whence $F = A$. Therefore $AC$ would be a tangent of the incircle of the triangle $ABD$. But the lines $AB$ and $A... | Estonia | IMO Team Selection Contest I | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
0fcr | Problem:
Dado el polinomio $p(x)=x^{3}+B x^{2}+C x+D$, probar que si el cuadrado de una de sus raíces es igual al producto de las otras dos, entonces $B^{3} D=C^{3}$ | [
"Solution:\n\nLlamemos $r$, $s$ y $t$ a las tres raíces.\nEl polinomio lo podemos escribir así: $p(x)=(x-r)(x-s)(x-t)$.\nSi operamos\n$$\np(x)=x^{3}-(r+s+t) x^{2}+(r s+s t+t r) x-r s t\n$$\ne igualamos coeficientes, obtenemos las conocidas relaciones de Cardano-Vieta:\n$$\n\\left\\{\\begin{array}{r}\nr+s+t=-B \\\\\... | Spain | Fase Local | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
03mg | A bookshelf contains $n$ volumes, labelled $1$ to $n$ in some order. The librarian wishes to put them in the correct order as follows. The librarian selects a volume that is too far to the right, say the volume with label $k$, takes it out, and inserts it so that it is in the $k$-th place. For example, if the bookshelf... | [
"(a) If $t_k$ is the number of times that volume $k$ is selected, then we have $t_k \\le 1 + (t_1 + t_2 + \\dots + t_{k-1})$. This is because volume $k$ must move to the right between selections, which means some volume was placed to its left. The only way that can happen is if a lower-numbered volume was selected.... | Canada | Kanada 2012 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | English, French | proof and answer | 2^{n-1} - 1 | |
03bn | Consider acute $\triangle ABC$ with altitudes $AA_1, BB_1$ and $CC_1$ ($A_1 \in BC, B_1 \in AC, C_1 \in AB$). A point $C'$ on the extension of $B_1A_1$ beyond $A_1$ is such that $A_1C' = B_1C_1$. Analogously, a point $B'$ on the extension of $A_1C_1$ beyond $C_1$ is such that $C_1B' = A_1B_1$ and point $A'$ on the exte... | [
"Let $O$ be circumcenter of $\\triangle ABC$ and $S$ be its symmetric point with respect to $AB$. Then $SOCH$ is a parallelogram. $SH \\perp A_1B_1$ and $SH = R$ (1). Denote by $\\rho$ the inradius of triangle $A_1B_1C_1$, and by $q$ its semi perimeter. If $T$ is the orthogonal projection of $S$ to $A_1B_1$ then $H... | Bulgaria | Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
080r | Problem:
Determinare tutte le soluzioni $(a, b)$, con $a, b$ interi relativi, dell'equazione
$$
a^3 + b^3 = 91
$$ | [] | Italy | Progetto Olimpiadi di Matematica 2000 GARA di SECONDO LIVELLO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (4, 3), (3, 4), (6, -5), (-5, 6) | |
00oz | A train consists of $2010$ wagons containing gold coins, all of the same shape. Any two coins have equal weight provided that they are in the same wagon, and differ in weight if they are in different ones. The weight of a coin is one of the positive reals $m_1 < m_2 < \dots < m_{2010}$. Each wagon is marked by a label ... | [
"We will prove that a single measurement is sufficient to do the job. Let $a_1 < a_2 < \\dots < a_n$ and $b_1 > b_2 > \\dots > b_n$, $n \\ge 2$ be arbitrary real numbers. For each permutation $\\pi$ of the set $\\{1, 2, \\dots, n\\}$ define\n$$\nZ(\\pi) = a_1 b_{\\pi(1)} + a_2 b_{\\pi(2)} + \\dots + a_n b_{\\pi(n)}... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | 1 | |
08m8 | Problem:
Determine all positive integer numbers $k$ for which the numbers $k+9$ are perfect squares and the only prime factors of $k$ are $2$ and $3$. | [
"Solution:\n\nWe have an integer $x$ such that\n$$\nx^2 = k + 9\n$$\nwhere $k = 2^a 3^b$, $a, b \\geq 0$, $a, b \\in \\mathbb{N}$.\nTherefore,\n$$\n(x-3)(x+3) = k.\n$$\nIf $b = 0$ then we have $k = 16$.\n\nIf $b > 0$ then we have $3 \\mid k + 9$. Hence, $3 \\mid x^2$ and $9 \\mid k$.\nTherefore, we have $b \\geq 2$... | JBMO | 2009 Shortlist JBMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 16, 27, 72, 216, 432, 2592 | |
0gs9 | Let $P(x)$ be a non-constant polynomial with real coefficients such that all of its roots are real numbers. Suppose that there exists a polynomial $Q(x)$ with real coefficients such that
$$
(P(x))^2 = P(Q(x))
$$
for all real numbers $x$. Prove that all roots of $P(x)$ are equal. | [
"Let $P(x) = A(x - r_1)^{d_1} \\cdots (x - r_k)^{d_k}$ where $r_1 < \\cdots < r_k$. It is easy to see that $Q(x)$ has degree $2$ and hence $Q(x) = a x^2 + b x + c$ for some real numbers $a$, $b$ and $c$. Then the given equality can be written as\n$$\nA^2 \\prod_{i=1}^{k} (x - r_i)^{2d_i} = A \\prod_{i=1}^{k} (a x^2... | Turkey | Team Selection Test for IMO 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induct... | English | proof only | null | |
040i | A non-negative number $m$ is called a *six match number*. If $m$ and the sum of its digits are both multiples of 6, find the number of the six match numbers less than 2012. | [
"Let $n = \\overline{d_1d_2d_3d_4} = 1000d_1 + 100d_2 + 10d_3 + d_4$, $d_1, d_2, d_3, d_4 \\in [0, 1, 2, \\dots, 9]$, and $S(n) = d_1 + d_2 + d_3 + d_4$.\nMatch the non-negative multiples of 6 less than 2000 into 167 pairs $(x, y)$, $x + y = 1998$, such that\n$(0, 1998), (6, 1992), (12, 1986), \\dots, (996, 1002).$... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 168 | |
0lb5 | Let $n$ be an integer larger than $1$. There are $n$ students sitting at a round table, each is given a number of candies (some students may receive no candies) and the sum of the candies of all students is a multiple of $n$. The students transfer the candies to each other in the following way:
With the candies given ... | [] | Vietnam | IMO2011 Selection | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
06g8 | Let $x_1$ be a positive real number and
$$
x_{n+1} = \sqrt{5x_n + 2\sqrt{x_n^2 + 1}} \quad \text{for } n = 1, 2, 3, \dots
$$
Prove that among $x_1, x_2, \dots, x_{2011}$, there are at least 670 irrational numbers. | [
"We rewrite the recurrence relation as follows.\n$$\n\\begin{aligned}\nx_{n+1} &= \\sqrt{5}x_n + 2\\sqrt{x_n^2 + 1} \\\\\n\\Rightarrow \\quad (x_{n+1} - \\sqrt{5}x_n)^2 &= 4(x_n^2 + 1) \\\\\n\\Rightarrow \\quad x_n^2 - 2\\sqrt{5}x_{n+1}x_n + (x_{n+1}^2 - 4) &= 0.\n\\end{aligned}\n$$\nThis is a quadratic equation in... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0h0m | Numbers $1$, $2$, ..., $n$ are placed in a row in a certain order. We are allowed to do the following operation: take any two pairs of consecutive elements, that have no common elements, and exchange their positions. Is it possible to get a monotone sequence of elements after a finite number of steps if:
a) $n = 2009$... | [
"**Answer:** a) not always; b) always possible.\n\nIf we consider $5$ consecutive elements, then from combination $12345$ we can get the following:\n$$\n12345 \\rightarrow 14523 \\rightarrow 23514 \\rightarrow 51234 \\rightarrow 53412 \\rightarrow 12453.\n$$\nLast three permutations were cyclically rearranged, so, ... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | a) not always; b) always possible. | |
0alc | Problem:
Find the value of $\log_{2}\left[2^{3} 4^{4} 8^{5} \cdots\left(2^{20}\right)^{22}\right]$.
(a) 3290
(b) 3500
(c) 3710
(d) 4172 | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | a | |
09w8 | On a circle with centre $M$ there are three distinct points $A$, $B$, and $C$ such that $|AB| = |BC|$. The point $D$ lies inside the circle in such a way that $\triangle BCD$ is isosceles. The second intersection point of $AD$ and the circle is called $F$. Prove that $|FD| = |FM|$. | [
"We will prove that $|FD| = |FC|$ and $|FC| = |FM|$, which proves the statement.\n\nIn the cyclic quadrilateral $ABCF$, we have $\\angle BCF = 180^\\circ - \\angle BAF$. As $|AB| = |BC| = |BD|$ we also have $\\angle BAF = \\angle BAD = \\angle ADB$, hence $\\angle BDF = 180^\\circ - \\angle ADB = 180^\\circ - \\ang... | Netherlands | BxMO Team Selection Test, March 2020 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fmy | Let $x, y, z$ be positive numbers such that $x^2y^2 + y^2z^2 + z^2x^2 = 6xyz$. Prove that
$$
\sqrt{\frac{x}{x + yz}} + \sqrt{\frac{y}{y + zx}} + \sqrt{\frac{z}{z + xy}} \ge \sqrt{3}
$$ | [
"Let $f : (-1, +\\infty) \\rightarrow \\mathbb{R}$ be the function defined by $f(x) = \\frac{1}{\\sqrt{1+x}}$. Since $f'(x) = -\\frac{1}{2(1+x)^{3/2}} < 0$ and $f''(x) = \\frac{3}{4(1+x)^2\\sqrt{1+x}} > 0$, then $f$ is convex. Applying Jensen's inequality to the function $f$ with $a_1 = \\frac{yz}{x}$, $a_2 = \\fra... | Spain | Mediterranean Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | Spanish | proof only | null | |
04jt | Let $N$ be a positive integer. In each square of a $N \times N$ array initially there is a zero. In each move it is allowed to choose a row or a column, erase all the numbers in it and then write numbers from $1$ to $N$ in arbitrary order. What is the maximal possible value of the sum of all numbers in the array? | [
"Let $m = n = N$ and let us assume that after some number of moves number $i$ appears $A_i$ times in the array ($1 \\le i \\le N$). If we consider the squares in which there is number $N$ to be black, and all other squares to be white, then the above statement implies\n$$\nA_N \\le N^2 - (N-1)^2.\n$$\n\nMore genera... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (4N^3 + 3N^2 - N)/6 | |
01xm | Does there exist a function $f(x): \mathbb{R} \to \mathbb{R}$ satisfying the equality
$$
f(|x|) + |f(x)| = x
$$
for all real numbers $x$? | [
"Answer: no, such function doesn't exist."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | No; such a function does not exist. | |
07p5 | Suppose $a_1, \dots, a_n > 0$, where $n > 1$ and $\sum_{i=1}^n a_i = 1$. For $i = 1, 2, \dots, n$, let $b_i = a_i^2 / \sum_{j=1}^n a_j^2$. Prove that
$$
\sum_{i=1}^{n} \frac{a_i}{1 - a_i} \le \sum_{i=1}^{n} \frac{b_i}{1 - b_i}.
$$
When does equality occur? | [
"Without loss of generality, we assume that $a_1 \\ge a_2 \\ge \\dots \\ge a_n$, so also $b_1 \\ge b_2 \\ge \\dots \\ge b_n$ and $0 < a_i, b_i < 1$. Let $S = \\sum_{i=1}^n a_i^2$, then $b_i = a_i^2/S$. Note that $S = S \\cdot \\sum_{i=1}^n a_i$ and $\\sum_{i=1}^n b_i = 1$. For $1 \\le k \\le n$ let\n$$\nD_k = \\sum... | Ireland | Irska 2014 | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | Equality holds if and only if a_1 = a_2 = ... = a_n = 1/n. | |
09n1 | Given integers $A$ and $B$ such that the sum of the digits of $A$ is $60$ and the sum of the digits of $B$ is $2024$, determine the least value of the sum of the digits of $A + B$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 5 | |
0hz6 | Problem:
Stacy has $d$ dollars. She enters a mall with 10 shops and a lottery stall. First she goes to the lottery and her money is doubled, then she goes into the first shop and spends 1024 dollars. After that she alternates playing the lottery and getting her money doubled (Stacy always wins) then going into a new s... | [
"Solution:\n\nWork backwards. Before going into the last shop she had $1024$, before the lottery she had $512$, then $1536$, $768$, $2304$, $1152$, and so on. We can easily prove by induction that if she ran out of money after $n$ shops, $0 \\leq n \\leq 10$, she must have started with $1024 - 2^{10-n}$ dollars. Th... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1023 | |
04vj | Consider the sequence $(a_n)_{n=1}^\infty$ defined as follows:
$$
a_1 = 3 \quad a_n = a_1a_2a_3 \dots a_{n-1} - 1 \quad \text{for all } n \ge 2.
$$
*Prove that there exist*
a) infinitely many primes dividing at least one member of this sequence;
b) infinitely many primes dividing no member of this sequence. | [
"a) By mathematical induction, we first prove that $a_n \\ge 2$ for every $n$. For $n=1$ and $n=2$ this is true because $a_1 = 3$ and $a_2 = 2$. Now suppose that for some $n \\ge 3$ the inequality $a_k \\ge 2$ holds for every $k < n$. Then we have $a_n = a_1a_2a_3\\dots a_{n-1} - 1 \\ge a_1a_2 - 1 = 5$, so indeed $... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0jrd | Problem:
Let $N = 30^{2015}$. Find the number of ordered 4-tuples of integers $(A, B, C, D) \in \{1, 2, \ldots, N\}^4$ (not necessarily distinct) such that for every integer $n$, $A n^3 + B n^2 + 2 C n + D$ is divisible by $N$. | [
"Solution:\nAnswer: 24\n\nNote that $n^{0} = \\binom{n}{0}$, $n^{1} = \\binom{n}{1}$, $n^{2} = 2\\binom{n}{2} + \\binom{n}{1}$, $n^{3} = 6\\binom{n}{3} + 6\\binom{n}{2} + \\binom{n}{1}$ (generally see http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind). Thus the polynomial rewrites as\n$$\n6A \\binom{... | United States | HMMT February 2015 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof and answer | 24 | |
0d9k | Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BH_b$, $CH_c$ intersecting at orthocenter $H$ ($H_b \in AC$, $H_c \in AB$). $H_b H_c$ meets $BC$ at $P$. Let $N$ be the midpoint of $AH$, $L$ be the orthogonal projection of $O$ on the symmedian with respect to angle $A$ of triangle $ABC$. Prove tha... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellane... | English | proof only | null | |
0a8s | Problem:
Let $f_{1}=0$, $f_{2}=1$, and $f_{n+2}=f_{n+1}+f_{n}$, for $n=1,2, \ldots$, be the Fibonacci sequence. Show that there exists a strictly increasing infinite arithmetic sequence none of whose numbers belongs to the Fibonacci sequence. [A sequence is arithmetic, if the difference of any of its consecutive terms... | [
"Solution:\n\nThe Fibonacci sequence modulo any integer $n>1$ is periodic. (Pairs of residues are a finite set, so some pair appears twice in the sequence, and the sequence from the second appearance of the pair onwards is a copy of the sequence from the first pair onwards.) There are integers for which the Fibonac... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 18 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
00xc | Problem:
Let $a \leq b \leq c$ be the sides of a right triangle, and let $2p$ be its perimeter. Show that
$$
p(p-c) = (p-a)(p-b) = S
$$
where $S$ is the area of the triangle. | [
"Solution:\n\nBy straightforward computation, we find:\n\n$$\n\\begin{aligned}\n& p(p-c) = \\frac{1}{4}\\left((a+b)^2 - c^2\\right) = \\frac{ab}{2} = S, \\\\\n& (p-a)(p-b) = \\frac{1}{4}\\left(c^2 - (a-b)^2\\right) = \\frac{ab}{2} = S.\n\\end{aligned}"
] | Baltic Way | Baltic Way 1992 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0fuy | Problem:
Betrachte einen See mit zwei Inseln darin und sieben Städten am Ufer. Die Inseln und Städte nennen wir im Folgenden kurz Orte. Zwischen genau den folgenden Paaren von Orten besteht eine Schiffsverbindung:
(i) zwischen den beiden Inseln,
(ii) zwischen jeder Stadt und jeder Insel,
(iii) zwischen zwei Städten ge... | [
"Solution:\n\nWir formulieren die Aufgabe um in die Sprache der Graphentheorie. Die Inseln werden durch Punkte $A, B$ repräsentiert, die sieben Städte durch Punkte $S_{1}, \\ldots, S_{7}$. Zwei Städte sind benachbart, falls sie aufeinanderfolgende Indizes $(\\bmod 7)$ haben. Je zwei dieser Punkte sind mit einer Kan... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0b7v | Let $n$ be an integer, $n \ge 2$. Find the remainder of the division of the number $n(n+1)(n+2)$ by $n-1$. | [
"We notice that $n(n+1)(n+2) = (n-1+1)(n-1+2)(n-1+3) = (n-1)^3 + 6(n-1)^2 + 11(n-1) + 6$. If $n-1 > 6$, the residue is $6$.\n\nIf $n = 2, 3, 4, 7$, the residue is $0$. If $n = 5$, the residue is $2$. If $n = 6$, the residue is $1$."
] | Romania | Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | n(n+1)(n+2) ≡ 6 (mod n−1). Hence the remainder is 6 for n ≥ 8; for n = 2,3,4,7 it is 0; for n = 5 it is 2; for n = 6 it is 1. | |
0gyv | In the rectangular trapezoid $ABCD$ the side $CD$ is perpendicular to the bases. Circle with diameter $AB$ intersects $AD$ at $A$ and $P$, the tangent to this circle at the point $P$ intersects line $CD$ at $M$. From the point $M$ the line $l$ is drawn, which is tangent to our circle at the point $Q$. Prove that the li... | [
"Let $N$ be the point of intersection of $BQ$ and $CD$, $L$ be the point of intersection of $BQ$ and $AD$ and $O$ be the midpoint of $AB$. Denote the radius of our circle by $r$. We have that (fig.22)\n$$\\angle NBC = \\angle NLD = \\frac{1}{2}(\\angle AQ - \\angle BP) = \\frac{1}{2}(\\angle AOQ - \\angle BOP) = \\... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05xl | Problem:
Soit $P(X)$ un polynôme à coefficients réels unitaire de degré $2022$. Emile joue au jeu suivant : il écrit le polynôme $P(X)$ au tableau et à chaque étape, si le polynôme $f(X)$ est écrit au tableau, Emile peut le remplacer par :
- Le polynôme $f(X)+c$, pour $c$ un réel de son choix, ou
- le polynôme $P(f(X)... | [
"Solution:\n\nOn dit qu'un entier $n$ fonctionne si, quel que soit le polynôme $P$ initial, Emile peut trouver une suite d'opérations telles que le polynôme obtenu ait exactement $n$ racines réelles distinctes.\n\nCommençons par faire la remarque simple qu'à chaque étape, quoique choisisse Emile, le polynôme écrit ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | All even n and all odd n at least 2021. | |
0dyn | a. Show that there is no positive integer $n$ such that the sum of the digits of $10^n + 9n$ is divisible by $2007$.
b. Find at least one positive integer $n$ such that the sum of the digits of $10^n + 9n$ equals $2008$. | [
"a. An integer is divisible by $9$ if and only if the sum of its digits is divisible by $9$. Assume that the sum of the digits of $10^n + 9n$ is divisible by $2007$. Since $2007$ is a multiple of $9$, the sum of the digits of $10^n + 9n$ must be divisible by $9$. This implies that $10^n + 9n$ is divisible by $9$, w... | Slovenia | Slovenija 2008 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Part (a): No such positive integer exists. Part (b): n = 111...111 (223 ones). | |
07br | A non-empty set $S$ of positive real numbers is called **powerful** if for any two distinct elements of it like $a$ and $b$, at least one of the numbers $a^b$ or $b^a$ is an element of $S$.
a) Present an example of a powerful set having four elements.
b) Prove that a finite powerful set cannot have more than four ele... | [
"a) $\\{1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{16}\\}$ is an example of a powerful set with four elements. (Part b shows that this is the unique powerful set with four elements.)\n\nb) First we prove a lemma.\n\n**Lemma 1.** A *finite powerful set* $S$ *can not have an element greater than one and an element less... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | {1, 1/2, 1/4, 1/16}; the maximum size of a finite powerful set is 4 | |
02nn | Let $ABCD$ be a parallelogram and $\Gamma$ the circumcircle of the triangle $ABD$. Lines $BC$ and $CD$ meet $\Gamma$ at $E \ne B$ and $F \ne D$ respectively. Prove the circumcenter of the triangle $CEF$ lies on $\Gamma$. | [
"Consider all angles oriented and modulo $180^\\circ$. Let $O$ be the center of the circle. In the cyclic pentagon $ABEDF$, $\\angle EBF = \\angle EDF = \\angle EDC = \\angle CED + \\angle DCE = \\angle BED + \\angle DCE = \\angle BAD + \\angle DCE = 2\\angle DCE$. This means that if $M$ is the midpoint of the arc ... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle"
] | null | proof only | null | |
00qs | Let $m$, $n$ be positive integers and $a$, $b$ be positive real numbers different from $1$. Suppose that $m > n$ and $\frac{a^{m+1}-1}{a^m-1} = \frac{b^{n+1}-1}{b^n-1} = c$. Show that $a^m c^n > b^n c^m$. | [
"Note that\n$$\n\\frac{b^{n+1}-1}{b^n-1} = \\frac{b^n + b^{n-1} + \\dots + 1}{b^{n-1} + \\dots + 1}\n$$\nHence\n$$\nc(b^{n-1} + \\dots + 1) = b^n + b^{n-1} + \\dots + 1 = b(b^{n-1} + \\dots + 1) + 1 \\Rightarrow c > b.\n$$\nWe also have\n$$\nc(b^{n-1} + \\dots + 1) = b^n + b^{n-1} + \\dots + 1 \\Rightarrow (c-1)(b^... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof only | null | |
0c3l | Problem:
Pentru orice număr natural nenul $n$ şi orice matrice coloană
$$
\mathbf{X} = \left(\begin{array}{c}
x_{1} \\
x_{2} \\
\vdots \\
x_{n}
\end{array}\right) \in \mathcal{M}_{n, 1}(\mathbb{Z})
$$
notăm cu $\delta(\mathbf{X})$ cel mai mare divizor comun al numerelor $x_{1}, x_{2}, \ldots, x_{n}$. Fie $n \in \mathb... | [
"Solution:\n\nArătăm că (a) implică (b). Fie $\\mathbf{B}=\\left(b_{i j}\\right) \\in \\mathcal{M}_{n}(\\mathbb{Z})$, fie $\\mathbf{X}=\\left(x_{j}\\right) \\in \\mathcal{M}_{n, 1}(\\mathbb{Z})$ şi fie $\\mathbf{B X}=\\left(y_{i}\\right) \\in \\mathcal{M}_{n, 1}(\\mathbb{Z})$. Cum $y_{i}=\\sum_{j=1}^{n} b_{i j} x_{... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0398 | Find all positive integers $m$ such that
$$
\frac{2^m \alpha^m - (\alpha + \beta)^m - (\alpha - \beta)^m}{3\alpha^2 + \beta^2}
$$
is an integer for all integer values of $\alpha, \beta$ with $\alpha\beta \neq 0$. | [] | Bulgaria | Second selection test for IMO 2007, Vietnam | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | All positive integers m with m ≡ 1 or 5 (mod 6). | |
0a6v | Problem:
Alice and Bob are playing a game. First, Alice chooses a partition $\mathcal{C}$ of the positive integers into a (not necessarily finite) set of sets, such that each positive integer is in exactly one of the sets in $\mathcal{C}$. Then Bob does the following operation a finite number of times.
Choose a set $... | [
"Solution:\n\nWe will use the following notation. If $S \\subseteq \\mathbb{Z}^{+}$ is a subset of the positive integers and $D$ is defined as in the problem statement, define $d(S) = D \\backslash S$. In addition, the $d$-sequence of $S$, written $\\{d^{S}\\}_{n \\in \\mathbb{Z}^{+}}$, is defined recursively as $d... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Alice has a winning strategy. | |
0iof | Problem:
Determine with proof, a simple closed form expression for
$$
\sum_{d \mid n} \phi(d) \tau\left(\frac{n}{d}\right)
$$ | [
"Solution:\nWe claim the series reduces to $\\sigma(n)$. The series counts the ordered triples $(d, x, y)$ with $d \\mid n;\\ x \\mid d;\\ 0 < y \\leq n/d;$ and $(y, n/d) = 1$. To see this, write\n$$\n\\sum_{d \\mid n} \\phi(d) \\tau\\left(\\frac{n}{d}\\right) = \\sum_{d' \\mid n} \\phi\\left(\\frac{n}{d'}\\right) ... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | σ(n) | |
07zu | Problem:
Dimostrare che un pentagono inscritto in una circonferenza e tale che ogni sua diagonale sia parallela ad un lato, è necessariamente regolare. | [] | Italy | Italy Febbraio Contest | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0gzr | The point $P$ lies inside triangle $ABC$. Denote by $O_A, O_B, O_C$ the circumcenters of triangles $PBC, PAC, PAB$ respectively. Let $O_P$ be the circumcenter of triangle $O_A O_B O_C$. Prove that the point $P$ satisfies the condition $O_P = P$ if $P$ is the orthocenter of triangle $ABC$. | [
"Let $A' = PA \\cap O_B O_C$, $B' = PB \\cap O_A O_C$, $C' = PC \\cap O_B O_A$. $O_A O_C$ is a perpendicular bisector of $BP$, thus $B'$ is a midpoint of $BP$. By analogy, $A', C'$ are midpoints of $PA, PC$ (Fig.07). If $P$ is a circumcenter of $\\triangle O_A O_B O_C$, then the perpendicular from $P$ to $O_A O_C$ ... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0g0r | Problem:
Bestimme alle natürlichen Zahlen $n$, sodass für beliebige reelle Zahlen $x_{1}, \ldots, x_{n}$ gilt:
$$
\left(\frac{x_{1}^{n}+\ldots+x_{n}^{n}}{n}-x_{1} \cdot \ldots \cdot x_{n}\right)\left(x_{1}+\ldots+x_{n}\right) \geq 0
$$ | [
"Solution:\n\nFür eins geht es trivialerweise. Die Ungleichung ist für drei erfüllt. Die Ungleichung lässt sich wie folgt faktorisieren:\n$$\n\\frac{1}{6}\\left(x_{1}+x_{2}+x_{3}\\right)^{2}\\left(\\left(x_{1}-x_{2}\\right)^{2}+\\left(x_{2}-x_{3}\\right)^{2}+\\left(x_{3}-x_{1}\\right)^{2}\\right) \\geq 0\n$$\nFür a... | Switzerland | IMO-Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | n = 1 and n = 3 | |
0594 | The orthocenter and the circumcenter of a non-equilateral triangle $ABC$ are $H$ and $O$, respectively. Let $D$ be the foot of the altitude dropped from the vertex $A$ of the triangle $ABC$. Prove that $\angle AHO = 90^\circ$ if and only if $\frac{AH}{HD} = 2$. | [
"Let $O'$ be the other endpoint of the diameter of the circumcircle of the triangle $ABC$ drawn from the vertex $A$ and let $H'$ be the reflection of the point $H$ through the line $BC$ (Fig. 35). Then $HD = H'D$, whence the condition $\\frac{AH}{HD} = 2$ is equivalent to the condition $\\frac{AH}{AH'} = \\frac{1}{... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0k0c | Problem:
The Fibonacci sequence is defined as follows: $F_{0}=0$, $F_{1}=1$, and $F_{n}=F_{n-1}+F_{n-2}$ for all integers $n \geq 2$. Find the smallest positive integer $m$ such that $F_{m} \equiv 0 \pmod{127}$ and $F_{m+1} \equiv 1 \pmod{127}$. | [
"Solution:\n\nFirst, note that $5$ is not a quadratic residue modulo $127$. We are looking for the period of the Fibonacci numbers $\\bmod\\ 127$. Let $p=127$. We work in $\\mathbb{F}_{p^{2}}$ for the remainder of this proof. Let $\\alpha$ and $\\beta$ be the roots of $x^{2}-x-1$. Then we know that $F_{n}=\\frac{\\... | United States | February 2017 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof and answer | 256 | |
0i4h | Problem:
A domino is a 1-by-2 or 2-by-1 rectangle. A domino tiling of a region of the plane is a way of covering it (and only it) completely by nonoverlapping dominoes. For instance, there is one domino tiling of a 2-by-1 rectangle and there are 2 tilings of a 2-by-2 rectangle (one consisting of two horizontal dominoes... | [
"Solution:\nThe number of tilings of a 2-by-$n$ rectangle is the $n$th Fibonacci number $F_{n}$, where $F_{0} = F_{1} = 1$ and $F_{n} = F_{n-1} + F_{n-2}$ for $n \\geq 2$. (This is not hard to show by induction.) The answer is $89$."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 89 | |
0g9y | 桌面上有 $n$ 張牌排成一圈。每張牌都有一面是黑的，另一面是白的。定義一次操作為：選擇一張黑面朝上的牌，將它和與它相鄰的兩張牌同時翻面。假設一開始時，只有 1 張黑面朝上的牌。試回答並證明：
(a) 若 $n = 2015$，是否能透過有限次操作讓所有牌都白面朝上？
(b) 若 $n = 2016$，是否能透過有限次操作讓所有牌都白面朝上？
Let $n$ cards are placed in a circle. Each card has a white side and a black side. On each move, you pick one card with black side up, flip it ove... | [
"(a) For $n = 2015$, you can make all cards white-side-up.\nLabel the cards from $1$ to $2015$ in order. WLOG assume the only black card initially is $2$. Consider the following moves (B=black, W=white):\n* $1$B, $2$W, $3$B\n* $1$B, $2$B, $3$W, $4$B\n* $1$B, $2$B, $3$B, $4$W, $5$B\n* ...\n* $1$B, $2$B, ..., $2013$B... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | For n = 2015: yes. For n = 2016: no. | |
087n | Problem:
In un'isola ci sono due tipi di abitanti: i cavalieri, che dicono sempre la verità, e i furfanti, che mentono sempre. Abbiamo incontrato su quest'isola un gruppo di quattro abitanti che, interrogati sulla loro identità, hanno risposto:
A: "C'è almeno un furfante tra noi."
B: "Ci sono al massimo due cavalieri ... | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. Le quattro affermazioni sono equivalenti a:\nA: \"C'è almeno un furfante tra noi.\"\nB: \"Ci sono almeno due furfanti tra noi.\"\nC: \"Ci sono almeno tre furfanti tra noi.\"\nD: \"Ci sono almeno quattro furfanti tra noi.\"\nSe ci sono $x$ furfanti, tutte e sole le prime ... | Italy | Olimpiadi di Matematica | [
"Discrete Mathematics > Logic"
] | null | MCQ | C | |
0ga2 | Find all ordered pairs $(a, b)$ of positive integers that satisfy $a > b$ and the equation $(a - b)^{ab} = a^b \cdot b^a$.
試求所有滿足 $(a-b)^{ab} = a^b \cdot b^a$ 且 $a > b$ 的正整數數對 $(a,b)$。 | [
"$(a, b) = (4, 2)$。\n\nLet $d$ be the greatest common divisor of $a$ and $b$, $a = dp$, $b = dq$, where $p, q$ are coprime positive integers and $p > q$. Substitute into the original equation:\n$$\n(d(p-q))^{d^2pq} = (dp)^{dq} \\cdot (dq)^{dp} \\quad \\Leftrightarrow \\quad (d(p-q))^{dpq} = (dp)^q \\cdot (dq)^p \\\... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (4, 2) | |
0b4d | Problem:
In triangle $ABC$ with orthocenter $H$, $AB = 13$, $BC = 21$ and $CA = 20$. The perpendicular bisector of $CH$ meets $BC$ at $P$ and lines $PH$ and $AB$ meet at $Q$. The line through $Q$ perpendicular to $PQ$ meets $AH$ at $X$. The length of $AX$ can be written in the form $p/q$, where $p$ and $q$ are relativ... | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 173 | |
06o8 | Let $n$ be a positive integer such that $1^3 + 2^3 + \dots + n^3$ is divisible by $n+3$. Find the greatest possible value of $n$. | [
"Answer: 15\nNote that\n$$\n1^3 + 2^3 + \\cdots + n^3 = \\frac{n^2(n+1)^2}{4}\n$$\n\nHence, for $n + 3$ to divide $1^3 + 2^3 + \\cdots + n^3$, it is necessary for $n^2(n + 1)^2$ to be a multiple of $n + 3$, or equivalently, by setting $m = n + 3$,\n$$\n0 \\equiv n^2(n+1)^2 = (m-3)^2(m-2)^2 \\equiv 36 \\pmod{m}.\n$$... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 15 | |
02x0 | Problem:
O retângulo $ABCD$ de medidas $AB=240~\mathrm{cm}$ e $BC=288~\mathrm{cm}$ representa um papel que será dobrado pelo segmento $EF$, onde $E$ pertence a $AD$ e $F$ pertence a $BC$, de modo que o ponto $C$ ficará sobre o ponto médio de $AB$.
a) Qual o comprimento de $CC^{\prime}$ ?
b) ... | [
"Solution:\nA operação de dobrar a folha de papel equivale a produzir uma reflexão ao longo da dobra. Para ver isso, perceba que segmentos correspondentes possuem o mesmo comprimento, por exemplo, $C^{\\prime}F = CF$ e $C^{\\prime}E = CE$. Assim, $EF$ divide o segmento $CC^{\\prime}$ ao meio e é ortogonal a ele.\n!... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | CC' = 312 cm, EF = 260 cm | |
09mq | Suppose $f(x)$ is a quadratic trinomial. If there exists another quadratic trinomial $g(x)$ such that $g(f(x)) = f(x) \cdot g(x)$ for all real $x$, we label $f(x)$ as *good* trinomial. How many positive integers are roots of a good trinomial and are less than or equal to 2025?
(Batzorig Undrakh) | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | 2024 | |
072x | The positive divisors $d_1, d_2, \dots, d_l$ of a natural number $n$ are arranged in the form
$$1 = d_1 < d_2 < \dots < d_l = n.$$
Suppose it is known that $d_1^2 + d_{15}^2 = d_{16}^2$. Find all possible values of $d_{17}$. | [
"We use the well known fact that given a triple $(a, b, c)$ of integers satisfying $a^2 + b^2 = c^2$, then one of $a, b$ is divisible by $4$; one of $a, b$ is divisible by $3$; and one of $a, b, c$ is divisible by $5$. Thus $4, 3$ and $5$ are divisors of $n$. Let us write\n$$n = 2^a 3^b 5^c t$$\nwhere $\\alpha \\ge... | India | Indija TS 2006 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 28 | |
09v6 | After breakfast, the sisters Anna and Birgit depart for school, each going to a different school. Their house is next to a bicycle path running between the two schools. Anna is cycling with a constant speed of $12$ km per hour and Birgit is walking in the opposite direction with a constant speed of $4$ km per hour. The... | [
"$42$"
] | Netherlands | Second Round, March 2019 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 42 | |
0f0p | Problem:
Players numbered $1$ to $1024$ play in a knock-out tournament. There are no draws, the winner of a match goes through to the next round and the loser is knocked-out, so that there are $512$ matches in the first round, $256$ in the second and so on. If $m$ plays $n$ and $m < n - 2$ then $m$ always wins. What i... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 21 | |
05oc | Problem:
On possède une liste infinie de cases, les cases étant numérotées par $1,2, \ldots$. Au départ, toutes les cases contiennent le nombre $1$. À chaque étape, on choisit un nombre $a \in \mathbb{N}^*$ tel que :
- soit toutes les cases numérotées par un multiple de $a$ contiennent $1$ auquel cas on remplace ces $... | [
"Solution:\n\nNous allons d'abord reformuler le problème. On se place dans $\\mathbb{N}^*$. Si $A$, $B$ sont deux ensembles (dans $\\mathbb{N}^*$), on note\n- $A \\oplus B := A \\cup B$ si $A \\cap B = \\emptyset$, sinon l'opération est interdite, et\n- $A \\ominus B := A \\setminus B$ si $B \\subset A$, sinon l'op... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0i7n | Problem:
A compact disc has the shape of a circle of diameter 5 inches with a 1-inch-diameter circular hole in the center. Assuming the capacity of the CD is proportional to its area, how many inches would need to be added to the outer diameter to double the capacity? | [
"Solution:\nDoubling the capacity is equivalent to doubling the area, which is initially $\\pi\\left[(5 / 2)^2 - (1 / 2)^2\\right] = 6\\pi$. Thus we want to achieve an area of $12\\pi$, so if the new diameter is $d$, we want $\\pi\\left[(d / 2)^2 - (1 / 2)^2\\right] = 12\\pi \\Rightarrow d = 7$. Thus we need to add... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles"
] | null | final answer only | 2 inches | |
00mn | Let $A$, $B$, $C$ and $D$ be four different points lying on a common circle in this order. Assume that the line segment $AB$ is the (only) longest side of the inscribed quadrilateral $ABCD$.
Prove that the inequality
$$
AB + BD > AC + CD
$$
holds.
(Karl Czakler) | [
"Let $S$ denote the common point of the diagonals, and let $a = AB$ and $c = CD$.\nSince $ABCD$ is an inscribed quadrilateral, triangles $ABS$ and $DCS$ are similar. It follows that numbers $r$ and $s$ must exist, such that $AS = sa$, $BS = ra$, $DS = sc$ and $CS = rc$ hold. The inequality under consideration can t... | Austria | 49th Austrian Mathematical Olympiad, National Competition (Final Round, part 2) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hlo | Problem:
Two players, Cat and Mouse, play the following game on a $4 \times 4$ checkerboard. Each player places a checker on a cell of the board (Cat goes first). Then, the two players take turns moving their checkers to an adjacent square, either vertically or horizontally (Cat again goes first). If, after either pla... | [
"Solution:\n\nMouse has a winning strategy. Color the cells of the checkerboard black and white alternately, in standard checkerboard fashion, so that adjacent cells are opposite colors. Whichever color Cat places his checker on, Mouse places his checker on a different cell of the same color. Then each time Cat mov... | United States | Berkeley Math Circle Monthly Contest | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Mouse has a winning strategy. | |
00su | If $a$, $b$, $c$ are positive real numbers such that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 3$, prove that
$$
\frac{a+b+c-1}{\sqrt{2}} \ge \frac{\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}}{3}
$$
When does equality hold? | [
"The inequality is equivalent to\n$$\n12(a + b + c - 1) \\ge \\sum_{\\text{cyc}} 4\\sqrt{2\\left(a + \\frac{b}{c}\\right)}.\n$$\nFrom AM-GM inequality we have\n$$\n\\sum_{\\text{cyc}} 2\\sqrt{2\\left(a+\\frac{b}{c}\\right)} \\le \\sum_{\\text{cyc}} \\left(2+a+\\frac{b}{c}\\right) = 6+a+b+c+\\frac{a}{b}+\\frac{b}{c}... | Balkan Mathematical Olympiad | BMO Short List | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds if and only if a = b = c = 1. | |
0hsy | Problem:
Prove that the functional equations
$$
\begin{aligned}
f(x+y) &= f(x) + f(y) \\
\text{and}\quad f(x+y+xy) &= f(x) + f(y) + f(xy) \quad (x, y \in \mathbb{R})
\end{aligned}
$$
are equivalent. | [
"Solution:\nLet us assume that $f(x+y) = f(x) + f(y)$ for all reals. In this case we trivially apply the equation to get $f(x+y+xy) = f(x+y) + f(xy) = f(x) + f(y) + f(xy)$. Hence the equivalence is proved in the first direction.\n\nNow let us assume that $f(x+y+xy) = f(x) + f(y) + f(xy)$ for all reals. Plugging in ... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
0f3j | Problem:
A rectangular box has sides $x < y < z$. Its perimeter is $p = 4(x + y + z)$, its surface area is $s = 2(xy + yz + zx)$ and its main diagonal has length $d = \sqrt{x^2 + y^2 + z^2}$. Show that $3x < (p/4 - \sqrt{d^2 - s/2})$ and $3z > (p/4 + \sqrt{d^2 - s/2})$. | [
"Solution:\nWe have $3(y - x)(z - x) > 0$, so $3x^2 + 3yz > 3xy + 3xz$. Hence $y^2 + z^2 + 4x^2 + 2yz - 4xy - 4xz > x^2 + y^2 + z^2 - xy - yz - xz$ or $(y + z - 2x)^2 > (d^2 - s/2)$. Hence $(x + y + z) > 3x + \\sqrt{d^2 - s/2}$. So $3x < p/4 - \\sqrt{d^2 - s/2}$.\n\nSimilarly, $3(z - x)(z - y) > 0$, so $x^2 + y^2 +... | Soviet Union | ASU | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
079w | Let $O$ be the circumcenter of triangle $ABC$. Points $A'$, $B'$, $C'$ lie on the segments $BC$, $CA$, $AB$ respectively such that the circumcircles of triangles $AB'C'$, $BC'A'$, and $CA'B'$ pass through $O$. Denote by $l_a$ the radical axis of the circle with center $B'$ and radius $B'C$ and the circle with center $C... | [
"*Proof.* We have $\\angle OB'C' = \\angle OAC' = 90^\\circ - \\angle C$ and $\\angle OB'A' = \\angle OCA' = 90^\\circ - \\angle A$. So $\\angle C'B'A' = \\angle B$. Compute the other angles similarly. This way it is proved that the triangles are similar and $O$ is the orthocenter of triangle $A'B'C'$. Now, the alt... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0iuy | Problem:
The Fibonacci sequence is the list of numbers that begins $1$, $2$, $3$, $5$, $8$, $13$ and continues with each subsequent number being the sum of the previous two.
Prove that when the first $n$ elements of the Fibonacci sequence are alternately added and subtracted, the result is an element of the sequence o... | [
"Solution:\n\nExpanding each number as the sum of the two previous numbers gives, for example,\n$$\n1-2+3-5+8-13=1-(1+1)+(1+2)-(2+3)+(3+5)-(5+8)\\text{.}\n$$\nNow, after removing parentheses, this is a telescoping series: each term subtracts out with its neighbor, so what's left is the last term of this expanded su... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0heu | Positive integers from $1$ to $100$ inclusive are written on the blackboard. Andrew wants to cross out some numbers in such a way, that the product of the remaining numbers is not divisible by $250$. What is the smallest number of numbers that he can cross? | [
"Since $250 = 2 \\cdot 5^3$, Andrew has to cross out from the product $1 \\cdot 2 \\cdots 100$ all numbers which are divisible by $5$, except for two numbers that are not divisible by $25$ (for example, we can leave $5$ and $10$). The resulting product satisfies the condition because it is not divisible by $5^3 = 1... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 18 | |
060y | Problem:
Soit $n \geqslant 1$ un entier et $x_{1}, \ldots, x_{n}$ des réels positifs. Montrer que
$$
\left(\frac{x_{1}}{1}+\frac{x_{2}}{2}+\ldots+\frac{x_{n}}{n}\right)\left(1 \cdot x_{1}+2 \cdot x_{2}+\ldots+n \cdot x_{n}\right) \leqslant \frac{(n+1)^{2}}{4 n}\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2}
$$
Quels sont le... | [
"Solution:\n\nL'astuce est d'appliquer une inégalité arithmético-géométrique au côté gauche de l'inégalité, mais avec des coefficients $\\sqrt{n}$ et $\\frac{1}{\\sqrt{n}}$ :\n$$\n\\begin{aligned}\n\\left(\\frac{x_{1}}{1}+\\frac{x_{2}}{2}+\\ldots+\\frac{x_{n}}{n}\\right)\\left(1 \\cdot x_{1}+2 \\cdot x_{2}+\\ldots+... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds for n = 1 trivially; for n ≥ 2, equality holds if and only if x_i = 0 for all i with 1 < i < n and x_1 = x_n. | |
0l5u | Problem:
Compute the number of ordered pairs $(m, n)$ of odd positive integers both less than $80$ such that
$$\gcd (4^{m} + 2^{m} + 1, 4^{n} + 2^{n} + 1) > 1.$$ | [
"Solution:\nFirst, we characterize all ordered pairs of general (not necessarily odd) positive integers $(m, n)$ such that $\\gcd (4^{m} + 2^{m} + 1, 4^{n} + 2^{n} + 1) > 1$. We claim that $(m, n)$ works if and only if either\n- $m$ and $n$ are both even, or\n- $\\nu_{3}(m) = \\nu_{3}(n)$.\n\nProof of necessity. Su... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 820 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.